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Pose Estimation Using Four Corresponding Points

Pose Estimation Using Four Corresponding Points. M.L. Liu and K.H. Wong, "Pose Estimation using Four Corresponding Points ", Pattern Recognition Letters, Volume 20, Number 1 January 1999, pp. 69-74. KH Wong. Introduction.

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Pose Estimation Using Four Corresponding Points

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  1. Pose Estimation Using Four Corresponding Points M.L. Liu and K.H. Wong, "Pose Estimation using Four Corresponding Points", Pattern Recognition Letters, Volume 20, Number 1 January 1999, pp. 69-74. KH Wong Pose estimation using 4 points v.4f

  2. Introduction • In this system, correspondences between four points on a rigid solid object and four points in an image are used to estimate the new position and orientation of that rigid object under full perspective projection. Synthetic and real data have been used to verify our method. Pose estimation using 4 points v.4f

  3. Problem definition • At time t, we have the model of the object A1,t, A2,t, A3,t, A4,t, f (focal length) • And corresponding image points a1,t, a2,t, a3,t, a4,t • We want to find the pose l1,t, l2,t, l3,t, l4,t • From l1,t, l2,t, l3,t, l4,t we can find the rotation and translation of the object with respect to the camera coordinates Pose estimation using 4 points v.4f

  4. Formulation • P is the center of perceptivity (camera center) • The vector PAn,t • passes through (unit vector un,t) passes through points an,t and An,t • The length is ln,t • Equation (1) and (2) Pose estimation using 4 points v.4f

  5. Error functions Pose estimation using 4 points v.4f

  6. Exercise 1: Newton’s method (An itervative method ) • An iterative method for finding the solution of a non-linear system • Exercise 1.Find sqrt(5), same as find the non-linear function. • f(x)=x2-5=0 • Taylor series (by definition) • f(x)=f(x0)+f’(x0)*(x-x0)=0 • f’(x0)=2*x0, so • f(x)=f(x0)+2*x0*(x-x0)=0 • First guess, x0=2. • f(x)=f(x0)+ f’(x0) *(x-x0)0 • 0  f(x0) + f’(x0) *(x-x0) • [0-f(x0)]/f’(x0)  (x-x0) • [0-(x02-5)]/2*x0 = x (x-x0) • [0-(x02-5)]/2*x0 = x • Take x0=2, [0-(22-5)]/2*2 = x • ¼= x • Since x (x-x0), • x=new guess, x0=old_guess • ¼  x-2, x  2.25 • That means the next guess is x  x2.25. • Exercise: Complete the steps to find the solution. • For your reference: sqrt(5)=2.2360679 (by calculator) • http://www.ugrad.math.ubc.ca/coursedoc/math100/notes/approx/newton.html Pose estimation V4h2

  7. Answer 1: Newton’s method • An iterative method for finding the solution of a non-linear system • Exercise 1.Find sqrt(5), same as find the non-linear function. sqrt(5)=2.2360679 (by calculator) • f(x)=x2-5=0 • Taylor series (by definition) • f(x)=f(x0)+f’(x0)*(x-x0)=0 • f’(x0)=2*x0, so • f(x)=f(x0)+2*x0*(x-x0)=0 • http://www.ugrad.math.ubc.ca/coursedoc/math100/notes/approx/newton.html Guess, x0=2.25 f(x)=f(x0)+2*x0*(x-x0)=0 f(x)=(x02-5)+2*x0*(x-x0)=0 f(x)=(x02-5)+2*x0*(x-x0)=0 (5.06-5)+2*2.25*(x-2.25)=0 0.06+4.5*(x-2.25)=0 X=((4.5*2.25)-0.06)/4.5 X=2.2366666 (temporally solution, but is good enough. ||Previous solution-current solution||2 =||2.25-2.2366666||2=0.013333 (small enough), continue if needed... Otherwise the solution is sqrt(5)=2.2366666. Pose estimation V4h2

  8. Using Taylor series http://www.fepress.org/files/math_primer_fe_taylor.pdf Pose estimation V4h2

  9. Netwon’s method: http://en.wikipedia.org/wiki/Newton%27s_method Pose estimation using 4 points v.4f

  10. The main idea of Newton's method • We saw this formula before: f(x)=f(x0)+f’(x0)*(x-x0)0 -----(i) • From f(x)=f(x0)+f’(x0)*(x-x0)0 • 0  f(x0)+f’(x0)*(x-x0) • 0 -f(x0)= f’(x0)*(x-x0) • [0 -f(x0)]/ f’(x0)=x=(x-x0) • We can compute x=[0 -f(x0)]/ f’(x0), then • Since x=(x-x0), so x=x0+ x • That means: Xnew_guess= x0(old_guess) + x Pose estimation V4h2

  11. The top 6 error functions Pose estimation using 4 points v.4f

  12. The extra error function Pose estimation using 4 points v.4f

  13. Stack up all error functions Pose estimation using 4 points v.4f

  14. http://en.wikibooks.org/wiki/Calculus/Taylor_series http://en.wikipedia.org/wiki/Taylor_series Stack up all 7 error functions Pose estimation using 4 points v.4f

  15. The iteration algorithm Pose estimation using 4 points v.4f

  16. Summary • A 4-point pose estimation algorithm is introduced. • The mathematical formulas of the algorithm are shown. • It is fast and accurate algorithm for pose estimation because no rotation matrix is used hence make it less complicated, stable and efficient. Pose estimation using 4 points v.4f

  17. Discussion • The error function g (or e4) corresponded to the square of the length of a vector. But the other errors e1,..,e4 are the squares of the squares of the difference between two vectors. They don’t seem to match. We can try to make gg^2 and see if it works better or not. My guess is it is more or less the same because near convergence they all become very small. Pose estimation using 4 points v.4f

  18. Appendix:Modified algorithmWarning: This modified Implementation is not very stable and accurate. • We prefer the Jacobean J to be a square matrix, so Inverse can be applied • Combine square-error function to make J a 4x4 matrix • e1=e12+e23 • e2=e14+e24 • e3=e34+e13 • e4=g Pose estimation using 4 points v.4f

  19. The terms of the Jacobean Pose estimation using 4 points v.4f

  20. Find g or e4 (the choice of points is not the same as the paper) Pose estimation using 4 points v.4f

  21. %*feb 2014*** jacob (a 4x4 matrix )for • %Liu M.L.and Wong K.H., "Pose Estimation • %Using Four Corresponding Points", • %Pattern Recognition Letters, Volume 20, • % Number 1 January 1999, pp. 69-74. • function four_point_algo_mlliu_khwong • clc,clear; • disp('test four_point_algo_mlliuTestJacobian'); • %assume they are all real • syms f l1 l2 l3 l4 cos_12 cos_13 cos_14 … • cos_23 cos_24 cos_34...\ • D12 D13 D14 D23 D24 D34 ...\ • te11 te12 te13 te21 te22 te23 real • %Vectors of all 6 combinations of 3D-points A1,A2,A3,A4 • %cos_12=cos(angle between unit vectors of PA1 and PA2),and(P=camera center) • dl = [l1,l2,l3,l4]'; • %---error functions of the 6 combinations, • %see equation 3 of liu+wong paper • e12=(l1^2+l2^2-2*l1*l2*cos_12-D12)^2; • e13=(l1^2+l3^2-2*l1*l3*cos_13-D13)^2; • e14=(l1^2+l4^2-2*l1*l4*cos_14-D14)^2; • e23=(l2^2+l3^2-2*l2*l3*cos_23-D23)^2; • e24=(l2^2+l4^2-2*l2*l4*cos_24-D24)^2; • e34=(l3^2+l4^2-2*l3*l4*cos_34-D34)^2; • e1=e12+e23 • e2=e14+e24 • e3=e34+e13 • % for g %-------------------------------------------- • g=l1*l2*l4*te11-l3*l1*l4*te12-l2*l3*l4*te13 ...\ • -l2*l1*l3*te21-l3*l1*l3*te22-l2*l3*l3*te23; • e4=g; • %------------------------------------------------- • E=[e1 e2 e3 e4 ]; • J = jacobian(E,dl); • disp('J ='), disp(J), size(J), Pose estimation using 4 points v.4f

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